1. Introduction
In this paper, all graphs are simple and connected. We refer to [1] for terms and notation that
are not defined in this paper.
Since graph labelings was first introduced by Rosa, various labeling
concepts have been introduced [2]. For example, cordial labeling, edge-friendly
labeling, harmonious labeling, felicitous labeling, odd harmonious
labeling and even harmonious labeling, semi-magic labeling, etc. Most
graph labeling methods can be traced to the method introduced by Rosa
[3] in 1967, or Graham
and Sloane [4] in
1980.
Let be a graph with
vertex set (or ) and edge set (or ). A labeling induces a vertex
labeling defined by , for each vertex . Here is the field of order 2.
For , let
: and : . The number is denoted by and is called the
edge-friendly index of .
An edge labeled by is called
an -edge
(under ). A vertex labeled by
is called an -vertex (under ). In this paper, an edge
labeling (or a -edge labeling) means an edge
labeling whose codomain is . An edge labeling of a graph is said to be
edge-friendly if .
Definition 1. The full edge-friendly index set
of , denoted by FEFI, is the set { :f is an edge-friendly labeling of
G}.
The concept of full edge-friendly index set was introduced in [5]. Since ,
So the edge-friendly index of of
is determined by the value of
.
Definition 2. Let be an edge labeling of a graph of order and size . We fix the sequence of vertices and the
sequence of edges of . The vector
is called an edge labeling
vector of , where
denotes the
Cartesian product of copies of
. Similarly, is called a vertex labeling
vector of .
For any (row) vector ,
denotes the transpose of .
Let be the incidence matrix of
a graph according to fixed
sequences of vertices and edges. Let be an edge-labeling of . It is easy to see that which is a column
vector of length over .
For a vector , the number of ’s in is denoted by , the Hamming weight of . Hence, if is an edge-friendly labeling of a graph
, then is . For convenience, we will identify
with . Hence .
Let be a graph and be fixed, . A one point
union of , , is the graph obtained from
the disjoint union of by
merging all into a single
vertex which is called the merged vertex or
core vertex. So there will be many different one
point unions of ’s.
If is a cycle of
order , then all one point
unions of , , are isomorphic. So we denote
the one point union of ,
, by for
. In this paper, we shall
study the full edge-friendly index set of the graph .
For , let . Let for , where . Let be the chosen vertex to be merged. We denote the
core vertex of the graph by
(or ). Note that is of order
and of size . Following is the one point union of , and , i.e., .
In the rest of the paper, we will use the notation defined above.
2. Some Extrema of
Edge-Friendly Indices
In this section, necessary and sufficient conditions that give the
extrema values of or are obtained. Following is
Lemma 2.1 of [5,6].
Lemma 1. Let be any edge labeling of a graph , then must be even.
Suppose there are odd numbers
among , where . Without loss of generality,
we may assume that
are odd and the others are even. Note that, means that there is no odd number;
means that there is no even
number. Now It is obvious that for any edge
labeling of . By Lemma 1 and the equation
above, we have Hence if is even for any
edge labeling of . A natural question is that whether
there is an edge-friendly labeling of attaining the lower bound and whether
there is an edge-friendly labeling of attaining the upper bound.
Lemma 2. Suppose there are odd numbers among . There is an edge-friendly
labeling of such that
if and only if
is even;
if and only if
is odd.
Proof. The necessity of (1) and (2) come from Eq. (3).
Note that is Eulerian. Let
be an Euler tour of . We label the edge of by 0 and 1 alternatively along . Clearly the induced label of each
vertex except the core is 1 and the label of the core is . Hence we have the
sufficiency of (1) and (2). 0
Obtaining the maximum value of is equivalent to obtaining the
minimum value of . 
Lemma 3. There is an edge-friendly labeling
of such that if and only if there are numbers among such that the sum of these
numbers is .
Necessity. Consider the subgraph induced by all -edges of under , . Since there is no 1-vertex, each
cycle of is entirely
edge-labelled by 1’s or entirely edge-labelled by 0’s. Hence is a one point union of some
subcycles of . Hence is also a one point union of some
subcycles of , namely is the one point union of subcycles. Since is edge-friendly, or
. Thus we
have the necessary condition by Remark 1.
[Sufficiency] Suppose there are numbers among such that the sum of such
numbers is . We label the edges of the corresponding
cycles by 0 and label the other edges of by 1. Thus this labeling is an
edge-friendly labeling and the labels of all vertices are 0’s. 
3. Full Edge-Friendly Index
Sets
The main results are given in this section. For a given one point
union of cycles , we fix the sequences of vertices and edges with
respect to the lexicographic order. Thus the incident matrix of is where , , and each
is a zero matrix of certain size,
.
We define some notation and recall some known results first. For
positive , let be the row vector of length whose entries are 1’s, and be the row vector of length whose entries are 0’s. Let and . For convenience, we let , , and be the empty rows.
Let . From (2), we
have , where
is defined in Section 2. So is odd if and only if is even. From now on, we shall use
.
For each edge labeling , where , , we have From (1) we
have For each possible value of the number of -vertices, we want to find an
edge-friendly labeling such that
.
We deal with some special cases first.
Theorem 1. Suppose all are even.
Suppose there are
numbers among such
that the sum of these numbers is , then
if is
odd;
if is
even.
Suppose the sum of any combination of integers is not equal to .
if is
odd;
if is
even.
Proof. For each and
, and , let . Thus, and . Hence
For , let . Thus,
. By (4)
Without loss of generality, we assume . Thus . Since
either or is at most . We may assume that and .
(1a) Suppose is odd.
Step 1: For a fixed , , let be an integer
such that .
Let be defined above, then (5) becomes
For a fixed , when runs through from to , the range of is an increasing arithmetic
sequence with common difference
from
to . Thus
when runs through from 1 to , the range of is an increasing arithmetic
sequence with common difference
from to
.
Step 2: Let be the Euler tour starts from the core
and travels the cycles , , , in order. Denote , where . In
this case .
There may have some ’s equal to
. Let ,
then .
Step 3: Now we swap the labels of
and ,
, where the indices are taken in modulo . For each case, the number of
1-vertices increases by 2 if or
is not incident with neither
nor ;
and may not change if is incident
with either or
for . The last case can occur
at most times. So the number
of 1-vertices increases by 2 at least times. Thus, the number of 1-vertices runs through each
even integers from 0 to under
the above swapping.
So we obtain all the possible values of the number of
1-vertices.
(1b) Suppose is even.
Perform Step 1 as Case (1a). Now (5) becomes
Similar to
Case (1a), when and run through all possible values, runs through , , …, .
Perform Steps 2 and 3 as Case (1a). We obtain that the number of
1-vertices runs through each even integers from 0 to .
So we obtain all the possible values of the number of
1-vertices.
By Lemma 3, for any edge-friendly labeling . Without loss of generality, we assume
that
and . Here .
(2a) Suppose is odd. Do the
same Step 1 as Case (1a) to obtain the edge labeling . Thus, we obtain that runs through , , …, .
Let be an Euler tour as in
Case (1a). In this case, . Constructing the
edge labeling as in Case (1a), we
get that , since .
Do the same Step 3 as in Case (1a). Here we obtain that the number of
1-vertices runs through each even integers from 2 to .
So we obtain all the possible values of the number of
1-vertices.
(2b) Suppose is even. By a
similar procedure and argument, we obtain all possible values of the
number of 1-vertices.
Hence this completes the proof. 
Theorem 2. Suppose all are odd.
Suppose there are
numbers among such
that the sum of these numbers is , then
.
Suppose the sum of any combination of integers is not equal to , then
.
Proof. Recall that and ; , , and are the empty rows (see the
beginning of this section).
For a fixed , , let be an integer such that . Let , then and . Hence .
For a fixed , , let be an integer such that . Let , then and . Hence .
Step 1: Let , where
and . Clearly
is edge-friendly. One may check that when and run through all possible values, runs through , , …, .
Step 2: For ,
we define , then and . Hence .
For , we define . Here and . Hence .
For each , , let . Note that,
. Clearly is edge-friendly. Suppose is even. We have .
One may check that when runs
through all even numbers from to
, runs through all even numbers
from down to .
Now we shall perform some steps similar to Steps 2 and 3 in the proof
of Theorem 1 to obtain the remaining possible values
of the number of 1-vertices.
Without loss of generality, we assume that . Here .
Let be the Euler tour starts from the core and travels the cycles , , , in order. By convention .
Step 3: Define . Now .
Step 4:
When , where . Note that , and
is odd. Swap the labels of and , . The number of 1-vertices
increases by 2 at least
times, since the first swapping increases by 2.
Next, swap the labels of and , . So the number of
1-vertices increases by 2 at least times.
By the same argument as Step 3 in the proof of Theorem 1, the number of 1-vertices increases by 2
at least times totally. Thus, the number of
1-vertices runs through each even integer from 0 to under the above swapping.
When , where (in this case cannot be 7). Note that , and
is odd.
Swap the labels of
and , . Next, swap the labels of
and , .
Totally, the number of 1-vertices increases by 2 at least
times. Thus, the number of 1-vertices runs through each even integer
from 0 to under the above
swapping.
When , where (in this case cannot be 8). Note that , and
is even.
Swap the labels of
and , . Next, swap the labels of
and , .
Totally, the number of 1-vertices increases by 2 at least times. Thus,
the number of 1-vertices runs through each even integer from 0 to under the above swapping.
When , where . Note that , and
is even.
Swap the labels of
and , . Next, swap the labels of
and , .
Totally, the number of 1-vertices increases by 2 at least times.
Thus, the number of 1-vertices runs through each even integer from 0 to
under the above
swapping.
Without loss of generality, we assume that and . Here . We do the same
procedure as Case 1. The only difference is . Hence the number of 1-vertices
at least runs through each even integer from 2 to under the above swapping.
This completes the proof. 
Example 1. Consider the graph . Here , , , and . Let be an Euler tour
of the graph, where . For , let , where .
The procedure of Steps 2 listed below:
The procedure of swapping (Steps 3 and 4) listed below start from
.
Example 2. Consider the graph . Here , , , and . Let be an Euler tour
of the graph, where . For
, let , where .
The procedure of Steps 3 and 4 listed below start from .
Lemma 4. Let and be two graphs with only one common
vertex . Suppose and be two edge-friendly labelings of
and , respectively. If , then there
is an edge-friendly labeling of
such that . If , then there is
an edge-friendly labeling of
such that .
Proof. Let be the
combined labeling of and . We obtain the lemma easily. 
Theorem 3. Let and be the order and the size of , respectively.
Suppose there are
numbers among such
that the sum of these numbers is , then .
Suppose there are no
numbers among such
that the sum of these numbers are , then .
Proof. Without loss of generality, we assume that are even and are odd, where . When or , we get the results using Theorem 1 and Theorem 2.
By Lemma 3, it suffices to find an
edge-friendly labeling of such that runs through all even numbers in
.
Let and .
Let and be orders of and , respectively. Note that is odd and .
By Theorem 1 there is an edge-friendly labeling of such that at least runs through all even
numbers of , and by
Theorem 2 there is an edge-friendly labeling of such that at least runs through all even
numbers of .
By Lemma 4, in the worst case, there is an
edge-friendly labeling of such that runs through all even numbers of
.
Now we only need to find an edge-friendly labeling of such that is
when is even, or when is odd, or else, is 2 (see 1⃝ and 2⃝ below).
Let be the Euler tour of starts from the core . Note that and .
When is even, . Define , then . Hence
When is odd, . Define , then . Hence
Suppose . Define . Thus only .
Hence .
Suppose , then and . Define
. Thus only .
Hence .
This completes the proof. 
We denote the graph by when . When , is called the Dutch
-windmill graph. By
Theorem 3, we have the following corollaries.
Corollary 1. Suppose and . Now, the order of is .
Suppose is odd, then
Suppose is even, then
Funding
The first author was partially supported by National Natural Science
Foundation of China (No: 12371344).
Financial Interests
The authors have no relevant financial or non-financial interests to
disclose.
Author
Contributions
All authors contributed to the study conception and design. Material
preparation and analysis were performed by all the authors. The first
draft of the manuscript was written by Zhen-Bin Gao and all authors
commented on previous versions of the manuscript. All authors read and
approved the final manuscript.